Suma a 2 numere mari mari c++

Codul l-am luat de aici:
https://www.geeksfor...-large-numbers/

Am gasit pe alt site si alta varianta, dar acolo codul mi s-a parut mult mai complex pentru mine.
Nu inteleg de ce IDE-ul are o problema cu <bits/stdc++.h>
Eu am sters .h si tot nu se compileaza.
Trebuie sa adaug vreun fisier ce se descarca de pe alt site?
Am VS 2022.
Eu am mai adaugat la cod #include <iostream>

Multumesc!

PS: Rog un moderator sa stearga din numele topicul unul din cuvintele "mari", ca apare de 2 ori.

// C++ program to find sum of two large numbers.
#include<bits/stdc++.h>
using namespace std;

// Function for finding sum of larger numbers
string findSum(string str1, string str2)
{
// Before proceeding further, make sure length
// of str2 is larger.
if (str1.length() > str2.length())
	 swap(str1, str2);

// Take an empty string for storing result
string str = "";

// Calculate length of both string
int n1 = str1.length(), n2 = str2.length();

// Reverse both of strings
reverse(str1.begin(), str1.end());
reverse(str2.begin(), str2.end());

int carry = 0;
for (int i=0; i<n1; i++)
{
	 // Do school mathematics, compute sum of
	 // current digits and carry
	 int sum = ((str1[i]-'0')+(str2[i]-'0')+carry);
	 str.push_back(sum%10 + '0');

	 // Calculate carry for next step
	 carry = sum/10;
}

// Add remaining digits of larger number
for (int i=n1; i<n2; i++)
{
	 int sum = ((str2[i]-'0')+carry);
	 str.push_back(sum%10 + '0');
	 carry = sum/10;
}

// Add remaining carry
if (carry)
	 str.push_back(carry+'0');

// reverse resultant string
reverse(str.begin(), str.end());

return str;
}

// Driver code
int main()
{
string str1 = "12";
string str2 = "198111";
cout << findSum(str1, str2);
return 0;
}

Edited by JustXXL, 10 May 2022 - 19:42.

Uite aici <https://www.geeksfor...g/bitsstdc-h-c/> ce e cu bits/stdc++.h, deci NU trebuie sa il incluzi. In schimb, trebuie sa incluzi ceea ce ai nevoie pentru ce folosesti in programel. Ca exemplu: #include <iostream> si #include <string>.

@sags, multumesc mult!
Acum se compileaza.